Three Dimensional Geometry Question 8
Question 8 - 2024 (27 Jan Shift 2)
The position vectors of the vertices $A, B$ and $C$ of a triangle are $2 \hat{i}-3 \hat{j}+3 \hat{k}, 2 \hat{i}+2 \hat{j}+3 \hat{k} \quad$ and $-\hat{i}+\hat{j}+3 \hat{k}$ respectively. Let $l$ denotes the length of the angle bisector $AD$ of $\angle BAC$ where $D$ is on the line segment $BC$, then $2 l^{2}$ equals :
(1) 49
(2) 42
(3) 50
(4) 45
Show Answer
Answer (4)
Solution
$AB=5$
$AC=5$
$\therefore D$ is midpoint of $BC$
$D\left(\frac{1}{2}, \frac{3}{2}, 3\right)$
$\therefore l=\sqrt{\left(2-\frac{1}{2}\right)^{2}+\left(-3-\frac{3}{2}\right)^{2}+(3-3)^{2}}$
$l=\sqrt{\frac{45}{2}}$
$\therefore 2 l^{2}=45$
